$\displaystyle ln(y)=$lim as x->0$\displaystyle ln(sinx/x + cosx/x)$
am I on the right track?
This is not true you cant bring the $\displaystyle \frac{1}{x} $ "into" your ln function. You want to write $\displaystyle \frac{\ln( \sin(x) + \cos(x))}{x} $